27107689 Engineering Economy Chapter 3x

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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Authored by Don Smith, Texas A&M University 2004 1

CHAPTER 3

COMBINING FACTORS

Mc GrawHill

ENGINEERING ECONOMY, Sixth Edition

by Blank and Tarquin





Learning Objectives

Shifted Series

Shifted Series and Single Amounts

Shifted Gradients

Decreasing Gradients

Spreadsheets





Section 3.1

Calculations For UniformSeries That Are Shifted





1. Uniform Series that are SHIFTED

A shifted series is one whose presentworth point in time is NOT t = 0.

Shifted either to the left of “0” or to the

right of t = “0”. Dealing with a uniform series:

The PW point is always one period to theleft of the first series value

No matter where the series falls on the time line.





Shifted Series

0 1 2 3 4 5 6 7 8

A = -$500/year

Consider:

P of this series is at t = 2 (P3 or F3)P3 = $500(P/A,i%,4) or, could refer to as F3

P0 = P3(P/F,i%,2) or, F3(P/F,i%,2)

P3 P0





Shifted Series: P and F

A = -$500/year

Consider:

F for this series is at t = 6

F6 = A(F/A,i%,4)

0 1 2 3 4 5 6 7 8

P3 P0

F at t = 6









3.2 Series with other single cash flows

It is common to find cash flows that arecombinations of series and other singlecash flows.

Solve for the series present worthvalues then move to t = 0

Solve for the PW at t = 0 for the singlecash flows

Add the equivalent PW’s at t = 0





3.2 Series with Other cash flows

Consider:

0 1 2 3 4 5 6 7 8

A = $500

F5 = -$400

F4 = $300

•Find the PW at t = 0 and FW at t = 8 for thiscash flow

i = 10%





3.2 The PW Points are:

F5 = -$400

F4 = $300

A = $500

0 1 2 3 4 5 6 7 8

i = 10%

t = 1 is the PW point for the $500 annuity;“n” = 3

1 2 3

C i h © Th M G Hill C i I P i i i d f d i di l





3.2 The PW Points are:

F5 = -$400

F4 = $300

A = $500

0 1 2 3 4 5 6 7 8

i = 10%

t = 1 is the PW point for the two othersingle cash flows

1 2 3

Back 4 periods

Back 5 Periods

C i ht © Th M G Hill C i I P i i i d f d ti di l





3.2 Write the Equivalence Statement

P = $500(P/A,10%,3)(P/F,10%,2)

+

$300(P/F,10%,4)

-

400(P/F,10%,5)

Substituting the factor values into theequivalence expression and solving….

C i ht © Th M G Hill C i I P i i i d f d ti di l





3.2 Substitute the factors and solve

P = $500( 2.4869 )( 0.8264 )

+

$300( 0.6830 )

-

400( 0.6209 )

=

$831.06

$1,027.58

$204.90

$248.36

Copyright © The McGraw Hill Companies Inc Permission required for reproduction or display





Section 3.3

Calculations for ShiftedGradients






3.3 The Linear Gradient - revisited

The Present Worth of an arithmeticgradient (linear gradient) is alwayslocated:

One period to the left of the first cash flowin the series ( “0” gradient cash flow) or,

Two periods to the left of the “1G” cash flow






3.3 Shifted Gradient

A Shifted Gradient is one whose presentvalue point is removed from time t = 0.

A Conventional Gradient is one whose

present worth point is t = 0.

Copyright © The McGraw-Hill Companies Inc Permission required for reproduction or display





3.3 Example of a Conventional Gradient

Consider:

……..Base Annuity ……..

Gradient Series

0 1 2 … n-1 n

This Represents a Conventional Gradient.

The present worth point is t = 0.






3.3 Example of a Shifted Gradient

Consider:

……..Base Annuity ……..

Gradient Series

0 1 2 … n-1 n

This Represents a Shifted Gradient.

The Present Worth Point for the

Base Annuity and the Gradientwould be here!




Copyright © The McGraw Hill Companies, Inc. Permission required for reproduction or display.


3.3 Shifted Gradient: Example

Given:

Base Annuity = $100

G = +$100

0 1 2 3 4 ……….. ……….. 9 10

Let C.F start at t = 3:

$500/ yr increasing by $100/yearthrough year 10; i = 10%; Find thePW at t = 0






3.3 Shifted Gradient: Example

PW of the Base Annuity

Base Annuity = $100

0 1 2 3 4 ……….. ……….. 9 10

P2 = $100( P/A,10%,8 ) = $100( 5.3349 ) = $533.49

Nannuity = 8 time periods

P0 = $533.49( P/F,10%,2 ) = $533.49( 0.8264 )

= $440.88






3.3 Gradient Present Worth

For the gradient component

0 1 2 3 4 ……….. ……….. 9 10

G = +$100

• PW of gradient is at t = 2:

•P2 = $100( P/G,10%,8 ) = $100( 16.0287 ) = $1,602.87

•P0 = $1,602.87( P/F,10%,2 ) = $1,602.87( 0.8264 )

• = $1,324.61




py g p , q p p y


3.3 Example: Final Solution

For the Base Annuity

P0 = $440.88

For the Linear Gradient

P0 = $1,324.61

Total Present Worth:

$440.88 + $1,324.61 = $1,765.49




py g p , q p p y


3.3 Shifted Geometric Gradient

Conventional Geometric Gradient

0 1 2 3 … … … n

A1

Present worth point is at t = 0 for a conventionalgeometric gradient!




py g p q p p y


3.3 Shifted Geometric Gradient

Conventional Geometric Gradient

0 1 2 3 … … … n

A1

Present worth point is at t = 2 for this example





3.3 Geometric Gradient Example

0 1 2 3 4 5 6 7 8

A = $700/yr

12% Increase/yr

i = 10%/year

A1 = $400 @ t = 5





3.3 Geometric Gradient Example

0 1 2 3 4 5 6 7 8

A = $700/yr

12% Increase/yr

i = 10%/year

PW point for thegradient

PW point for theannuity





3.3 The Gradient Amounts

t Base Amt

1 $400.00

2 $448.00

3 $501.76

4 $561.97

5

6

7

8

Present Worth of the Gradient at t = 4

P4

= $400{ P/A1,12%,10%,4 } = 1,494.70$

3.73674

P0 = $1,494.70( P/F,10%,4) = $1,494.70( 0. 6830 )

P0 = $1,020,88





3.3 The Annuity Present Worth

PW of the Annuity

0 1 2 3 4 5 6 7 8

i = 10%/year

P0 = $700(P/A,10%,4)

= $700( 3.1699 ) = $2,218.94

A = $700/yr





3.3 Total Present Worth

Geometric Gradient @ t =

P0 = $1,020,88

Annuity

P0 = $2,218.94

Total Present Worth”

$1,020.88 + $2,218.94

= $3,239.82







3.4 Shifted Decreasing Linear Gradients

Given the following shifted, decreasinggradient:

0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year

Find the Present Worth @ t = 0








0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year

P2 or, F2: Take back to t = 0

P0 here






0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year


P0 here

Base Annuity = $1,000





3.4 Time Periods Involved

0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year


P0 here

1 2 3 4 5

Dealing with n = 5.





3.4 Time Periods Involved

0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year

1 2 3 4 5

P2 = $1,000( P/A,10%,5 ) – 100( P/G,10%.5 )

$1,000 G = -$100/yr

P2= $1,000( 3.7908 ) - $100( 6.8618 ) = $3,104.62

P0 = $3,104.62( P/F,10%,2 ) = $3104.62( 0 .8264 ) = $2,565.65





Section 3.5

Spreadsheet Applications





3.5 Spreadsheet Applications

Assume Excel is the spreadsheet of choice

Instructors may vary on the degree of

emphasis placed on spreadsheet useStudent’s Goal:

Learn the Excel Financial Functions

Create your own spreadsheets to solve avariety of problems





3.5 NPV Function in Excel

NPV function is basic

Requires that all cell in the range sodefined have an entry.

The entry can be $0…but not blank! Incorrect results can be generated if one or more cells in the defined range isleft blank .

A “0” value must be entered.





3.5 Spreadsheets

It is assumed if an instructor desires toapply spreadsheets, he or she willprovide examples and go over each

example and the associated cellformulas.

See Appendix A for further details onExcel applications




End of Slide Set

Mc GrawHill

ENGINEERING ECONOMY, Sixth Edition

Blank and Tarquin

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27107689 Engineering Economy Chapter 3x